Lemma 29.50.1. Let $X$, $Y$ be schemes. Let $f : X \to Y$ be locally of finite type. Let $\eta \in Y$ be a generic point of an irreducible component of $Y$. The following are equivalent:

the set $f^{-1}(\{ \eta \} )$ is finite,

there exist affine opens $U_ i \subset X$, $i = 1, \ldots , n$ and $V \subset Y$ with $f(U_ i) \subset V$, $\eta \in V$ and $f^{-1}(\{ \eta \} ) \subset \bigcup U_ i$ such that each $f|_{U_ i} : U_ i \to V$ is finite.

If $f$ is quasi-separated, then these are also equivalent to

there exist affine opens $V \subset Y$, and $U \subset X$ with $f(U) \subset V$, $\eta \in V$ and $f^{-1}(\{ \eta \} ) \subset U$ such that $f|_ U : U \to V$ is finite.

If $f$ is quasi-compact and quasi-separated, then these are also equivalent to

there exists an affine open $V \subset Y$, $\eta \in V$ such that $f^{-1}(V) \to V$ is finite.

**Proof.**
The question is local on the base. Hence we may replace $Y$ by an affine neighbourhood of $\eta $, and we may and do assume throughout the proof below that $Y$ is affine, say $Y = \mathop{\mathrm{Spec}}(R)$.

It is clear that (2) implies (1). Assume that $f^{-1}(\{ \eta \} ) = \{ \xi _1, \ldots , \xi _ n\} $ is finite. Choose affine opens $U_ i \subset X$ with $\xi _ i \in U_ i$. By Algebra, Lemma 10.121.10 we see that after replacing $Y$ by a standard open in $Y$ each of the morphisms $U_ i \to Y$ is finite. In other words (2) holds.

It is clear that (3) implies (1). Assume $f$ is quasi-separated and (1). Write $f^{-1}(\{ \eta \} ) = \{ \xi _1, \ldots , \xi _ n\} $. There are no specializations among the $\xi _ i$ by Lemma 29.20.7. Since each $\xi _ i$ maps to the generic point $\eta $ of an irreducible component of $Y$, there cannot be a nontrivial specialization $\xi \leadsto \xi _ i$ in $X$ (since $\xi $ would map to $\eta $ as well). We conclude each $\xi _ i$ is a generic point of an irreducible component of $X$. Since $Y$ is affine and $f$ quasi-separated we see $X$ is quasi-separated. By Properties, Lemma 28.29.1 we can find an affine open $U \subset X$ containing each $\xi _ i$. By Algebra, Lemma 10.121.10 we see that after replacing $Y$ by a standard open in $Y$ the morphisms $U \to Y$ is finite. In other words (3) holds.

It is clear that (4) implies all of (1) – (3) with no further assumptions on $f$. Suppose that $f$ is quasi-compact and quasi-separated. We have to show that the equivalent conditions (1) – (3) imply (4). Let $U$, $V$ be as in (3). Replace $Y$ by $V$. Since $f$ is quasi-compact and $Y$ is quasi-compact (being affine) we see that $X$ is quasi-compact. Hence $Z = X \setminus U$ is quasi-compact, hence the morphism $f|_ Z : Z \to Y$ is quasi-compact. By construction of $Z$ we see that $\eta \not\in f(Z)$. Hence by Lemma 29.8.4 we see that there exists an affine open neighbourhood $V'$ of $\eta $ in $Y$ such that $f^{-1}(V') \cap Z = \emptyset $. Then we have $f^{-1}(V') \subset U$ and this means that $f^{-1}(V') \to V'$ is finite.
$\square$

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